TITLE: Probability of global epidemic
QUESTION [11 upvotes]: Consider $\mathbb{Z}^2$ as a graph, where each node has four neighbours. 4 signals are emitted from $(0,0)$ in each of four directions (1 per direction) . A node that receives one signal (or more) at a timestep will re-emit it along the 4 edges to its four neighbours at the next time step. A node that did not receive a signal at the previous timestep will not emit a signal irrespective of whether it earlier received a signal. There is a $50\%$ chance that a signal will be lost when travelling along a single edge between two neighbouring nodes.
A node that receives more then 1 signal acts the same as if it received only 1.
The emitting of a signal in each of the 4 directions are independent events.
What is the probability that the signal will sometime arrive at $(10^5,10^5)$?
Research: Simulations show: Yes. ~90%
What is the probability that the signal will sometime arrive at $(x,y)$ if a signal traveling along an edge dies with probability $0<p<1$?
What is the least p, for which the probability that N initial random live cells die out approaches 0, as N approaches infinity? Experiment shows p close to 0.2872.
In $\mathbb{Z}^1$, $p_{min}=0.6445...$, how to calculate this?
In $\mathbb{Z}^3$, $p_{min}=0.1775...$, how to calculate this?

REPLY [3 votes]: In the same vein as Didier's answer providing bounds on the extinction probability $q$, we can also obtain bounds on the transmission probability $p$ required for the probability of $N$ signals dying out to go to $0$ as $N\to\infty$, which is the probability required for the extinction probability $q$ not to be $1$.
Didier's equation for $q$ can be rewritten as $q=(1-(1-q)/2)^4$, which says that the signal goes extinct if all four signals die, either because they don't make it across the edge ($1/2$) or because they go extinct afterwards ($1-q$). Generalizing to $d$ dimensions and transmission probability $p$, this is
$$q=(1-p(1-q))^{2d}\;.$$
This equation always has a root at $1$. For sufficiently large $q$, it also has a second root in $[0,1]$, and the extinction probability is given by that root. The critical case in which the extinction probability becomes $1$ occurs when these two roots coincide. Differentiating with respect to $q$ and substituting $q=1$ yields the condition for $1$ to be a double root:
$$
\begin{eqnarray}
1&=&2dp(1-p(1-q))\;,
\\
1&=&2dp\;,
\\
p&=&\frac1{2d}\;.
\end{eqnarray}
$$
This is a lower bound, since in the case of confluent signals there are fewer signals to keep the fire burning. Your experiments seem to indicate that the bound becomes better with increasing dimension, which makes sense since the signals become less likely to collide.